hafind said:
First time poster, thanks for help
I wonder if low tide is caused by centrifugal force at noon time and high tide at midnight?
The centrifugal force pushes sea water toward the center of the Earth at day time, away from it at night time. So, a man weights more at day time than night time.
Am I have a point?
hafind said:
Thanks. I still don't understand.
The centrifugal force pushes everything on Earth away from the Sun. So the water facing the Sun will become low tide, the other side high tide. Moon gravity and Earth movement play part of the game.
Do you think centrifugal force has nothing to do with tide? Maybe we can put a gel ball on a turn table and see if the ball changes shape?
The too long, didn't read answer: The tides are not caused by centrifugal forces.You are using the term "centrifugal force" without apparently understanding it.
So let's back up a bit. First thing to note: The centrifugal force is an inertial force (or pseudo force or fictitious force).
The preferred frames of reference in Newtonian mechanics are the frames in which all three of Newton's laws of motion hold, the inertial frames of reference. There is no such thing as the centrifugal force in an inertial frame; all of the fictitious forces vanish in an inertial frame.
This means you can't attribute cause and effect to a fictitious force. What one can do is use fictitious forces as a convenient way of explaining some behavior. For example, it's far easier to explain why hurricanes rotate from the perspective of a frame rotating with the Earth than it is from the perspective of an inertial frame.
Back to the tides, the explanation from the perspective of an inertial frame is quite simple. Imagine two drops of water, one at the point on the surface of the Earth closest to the Moon, the other at the point on the surface of the Earth furthest from the Moon. Gravity makes both drops, along with the Earth as a whole, accelerate toward the Moon.
Gravitation is an inverse square law. The acceleration of the drop closest to the Moon is greater than is that of the Earth as a whole. That drop closest to the Moon is thus pulled away from center of the Earth. The acceleration of the Earth as a whole toward the Moon is greater than is that of the drop furthest from the Moon. The center of the Earth is thus pulled away from that drop furthest from the Moon. Another way to look at this: The drop furthest from the Moon also is pulled away from center of the Earth.
It's oftentimes convenient to explain the tides from the perspective of a frame of reference with it's origin at the center of the Earth. The Earth as a whole is accelerating toward the Moon (and the Sun, and also toward every other mass in the universe). This means an Earth-centered frame is not an inertial frame. It's an accelerating frame, and hence some kind of fictitious force is needed to explain behaviors in this frame. This is the accelerating frame fictitious force. From the perspective of this (accelerating) Earth-centered frame, the tidal forces look like this:
This fictitious force is *not* the centrifugal force. The accelerating frame fictitious force is a uniform force, one that is constant in direction and magnitude regardless of location. In comparison, the centrifugal force results from a rotating rather than accelerating frame. The centrifugal force is proportional to distance from the center of rotation and is directed away from the center of rotation.
Some oceanographers, and even some oceanography textbooks, try to explain the tides using the concept of rotating frames and the centrifugal force. This is, to be blunt, an erroneous explanation.
